UVA 12034-Race

Race

Time Limit: 1000MS Memory Limit: Unknown 64bit IO Format: %lld & %llu

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Description

Disky and Sooma, two of the biggest mega minds of Bangladesh went to a far country. They ate, coded and wandered around, even in their holidays. They passed several months in this way. But everything has an end. A holy person, Munsiji came into their life. Munsiji took them to derby (horse racing). Munsiji enjoyed the race, but as usual Disky and Sooma did their as usual task instead of passing some romantic moments. They were thinking- in how many ways a race can finish! Who knows, maybe this is their romance!

In a race there are n horses. You have to output the number of ways the race can finish. Note that, more than one horse may get the same position. For example, 2 horses can finish in 3 ways.

1. Both first
2. horse₁ first and horse₂ second
3. horse₂ first and horse₁ second

Input 

Input starts with an integer T ( 1000), denoting the number of test cases. Each case starts with a line containing an integer n ( 1n1000).

Output 

For each case, print the case number and the number of ways the race can finish. The result can be very large, print the result modulo 10056.

Sample Input 

3123

Sample Output 

Case 1: 1Case 2: 3Case 3: 13

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Source

Root :: AOAPC II: Beginning Algorithm Contests (Second Edition) (Rujia Liu) :: Chapter 10. Maths :: Examples
Root :: Prominent Problemsetters :: Md. Mahbubul Hasan
Root :: AOAPC I: Beginning Algorithm Contests -- Training Guide (Rujia Liu) :: Chapter 2. Mathematics :: Counting ::Exercises: Intermediate

小结：

    大概想法是递推，至于为什么是递推，因为题目分类就是递推（看到这里不要打我＝＝），对于递归的想法，我感觉，对于每种情况的分类讨论是解决这类问题的关键。

就拿这道题目来说，在现有的名次中插入新个体是这种题目的常见思路。

令dp[i][j]=i个人组成的所有的名次的种类数目，那么对应的每个新插入的个体，对应2种情况：(1)增加现有的名次总数 (2)加入现有的任意一个名次中。

然后就会有以下递推方程：dp[i][j]=i==j?dp[i-1][j-1]*j:(dp[i-1][j]+dp[i-1][j-1])*j;

以下是AC代码：

//111,112,121,211,122,212,221,123,132,213,231,312,321//当n的值为3的时候，排名的种数为13#include<stdio.h>#include<string.h>#include<math.h>#include<algorithm>using namespace std;const int mod=10056;int dp[1005][1005];void solve(){    int ans,n;    int t;    scanf("%d",&t);    for(int m=1;m<=t;m++)    {        scanf("%d",&n);        ans=0;        for(int i=1;i<=n;i++)        {            ans+=dp[n][i];            ans=ans%mod;        }        printf("Case %d: %d\n",m,ans);    }}int main(){    int n;    int mod=10056;    memset(dp,0,sizeof(dp));    dp[0][0]=1;    for(int i=1;i<=1000;i++)    {        for(int j=1;j<=i;j++)        {            dp[i][j]=i==j?dp[i-1][j-1]*j:(dp[i-1][j]+dp[i-1][j-1])*j;            dp[i][j]=dp[i][j]%mod;        }    }    solve();    return 0;}